Hi, $\displaystyle \int_{-1}^8 \frac{1}{x^{2/3}} \, dx$ How is this even possible since the integrand isn't even defined for values less than 0?...
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Originally Posted by sjmiller $\displaystyle \int_{-1}^8 \frac{1}{x^{2/3}} \, dx$ How is this even possible since the integrand isn't even defined for values less than 0? Why would you say that? For all $\displaystyle x\ne 0$ we have $\displaystyle \frac{1}{x^{2/3}}>0$.
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